I've been studying the recursion theorem in my introductory set theory course, and I've been given a homework about it. There was one exercise in particular that was, at first, kind of difficult to prove, but after attempting multiple things and realizing some others, it was actually really easy to solve.
Now, I'm having trouble understanding what this result means or how it could be useful/applied in other problems (this could mainly be because I'm yet to study higher level branches of mathematics and so I'm blind to these kind of things because of the lack of experience).
Here's the theorem in question:
Let $f\colon X\to X$ be a function and $A\subseteq X$. By the recursion theorem, there exists a unique function $h\colon\mathbb{N}\to\wp(X)$ such that $h(0)=A$ and for all $n\in\mathbb{N}$, $h(S(n))=h(n)\cup f[h(n)]$. Then $$\bigcap\{ Y\in\wp(X)|A\subseteq Y \land f[Y]\subseteq Y \}=\bigcup_{n\in\mathbb{N}}h(n)$$ This set is defined as the closure of $A$ under $f$.
So yeah, any help on how to interpret and use this theorem, and even before that, to really understand what this particular set does or means would be really appreciated. Thanks in advance.
$h(n)$ is what one would normally call $\bigcup\limits_{m=0}^nf^m[A]$. The theorem states that $\bigcup_{n\in\Bbb N}\bigcup\limits_{m=0}^nf^m[A]$ is the intersection of all the subsets which are $f$-stable and which contain $A$. If we put this together with the fact that $\bigcup_{n\in\Bbb N}\bigcup\limits_{m=0}^nf^m[A]$ is essentially equal to $\bigcup_{n\in\Bbb N}f^n[A]$, we are saying that an element is in all the $f$-stable subsets that contain $A$ if and only if either it is an element of $A$ or it can be recovered by applying $f$ a finite number of times to some element of $A$.